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anonymous

  • 5 years ago

How do I use the exponential shift to solve the linear differential equation y'' + 7y' +12y = 2(e^-3)(cos2x)

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  1. anonymous
    • 5 years ago
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    im looking to find the general solution y = yc +yp. so far i have solved the homogenous part of the equation yc. yc = Ae^-3x + Be^-4x. i cant solve for yp yet.

  2. anonymous
    • 5 years ago
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    ok

  3. anonymous
    • 5 years ago
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    Let me do it on paper first. I think it would be murder typing it out.

  4. anonymous
    • 5 years ago
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    haha thank you! any help will go a long way

  5. anonymous
    • 5 years ago
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    i'll continue to try draw something out of my mess of numbers and letters while i wait

  6. anonymous
    • 5 years ago
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    i also figured that the derivative operator 'D' helps so that the LHS = (D^2 + 7D +12)y

  7. anonymous
    • 5 years ago
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    You do mean e^{-3x}, don't you?

  8. anonymous
    • 5 years ago
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    yes! sorry typo

  9. anonymous
    • 5 years ago
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    When I get to factors that need expanding, I'm going to get Mathematica to do that (save time).

  10. anonymous
    • 5 years ago
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    ok no worries. if i dont follow il just let u know

  11. anonymous
    • 5 years ago
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    So what happened, your lecturer just dump theory on you with no demonstration?

  12. anonymous
    • 5 years ago
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    he demonstated with the d.e. y'' + 6y' + 13y = x(e^3x)(sin2x) however that takes y = imaginary(Y) and also takes into account the complex numbers. As my roots are reall -3, -4 and i take y=real(Y) instead, i dont know what yp to try

  13. anonymous
    • 5 years ago
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    he pretty much chalk vomits all over the board without going into the specifics of how he gets to places

  14. anonymous
    • 5 years ago
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    Not trained to teach...

  15. anonymous
    • 5 years ago
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  16. anonymous
    • 5 years ago
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    its really quite frustrating... luckily we are changing lecturers next week.. sorry for any hassle im causing you!

  17. anonymous
    • 5 years ago
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    just give me a fan point and i'll be happy

  18. anonymous
    • 5 years ago
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    wow thanks! im just gona go through it and check i understand.. haha ok sure how do i do that?

  19. anonymous
    • 5 years ago
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    just click my 'become a fan' link...if there isn't one there, refresh the page...should work then

  20. anonymous
    • 5 years ago
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    ps. is there a place to download mathmatica for free? ok right on it

  21. anonymous
    • 5 years ago
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    www.wolframalpha.com The guy who made mathematica made this site. It can do a lot of things like integration, solving differential equations, plotting, etc.

  22. anonymous
    • 5 years ago
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    fanned. thanks a million! will i still be able to ask help from you if i have any more troubles..

  23. anonymous
    • 5 years ago
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    this is wolfram's solution to your equation http://www.wolframalpha.com/input/?i=y%27%27-2y%27%2B5y+%3D+16x^3*e^%283x%29+++

  24. anonymous
    • 5 years ago
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    woops...no it's not...wait.

  25. anonymous
    • 5 years ago
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    haha! i was like crap.. i did notttttt get that

  26. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=y%27%27%2B7y%27%2B12y+%3D+2e^%28-3%29cos%282x%29

  27. anonymous
    • 5 years ago
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    haha ur awesome at pen to paper but im afraid you did the same typo as me last time. theres an 'x' in e^-3x. however the graphs dont look how i thought they would..

  28. anonymous
    • 5 years ago
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    graphs in the correct version*

  29. anonymous
    • 5 years ago
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    ah...typo...

  30. anonymous
    • 5 years ago
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    yep :) i thought it would have more of a decaying oscillation as it is overdamped.. is it because of the negative exponential? why?

  31. anonymous
    • 5 years ago
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    So, am I right in understanding you've finished deriving the solution?

  32. anonymous
    • 5 years ago
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    And you're reading off that and not Wolfram?

  33. anonymous
    • 5 years ago
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    sorry no. i was comparing my notes to the graphs of wolfram.. but! i havent read through the wolfram content yet.. just making observations. if you would like you can continue helping other students.. and im guessing when i need help again it will notify you?

  34. anonymous
    • 5 years ago
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    Yes, I get an e-mail if you respond to a thread I've been in. Are you fine to derive the rest?

  35. anonymous
    • 5 years ago
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    If an error's been made on my part, it's because it's after 1am where I am. If you need help in the next ~12-24hrs, I may not be around. You'd should try your luck with someone else.

  36. anonymous
    • 5 years ago
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    i should be ok for now thanks :) it takes me quite a while longer than you to come up with a solution

  37. anonymous
    • 5 years ago
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    yep me too. 1:22am. ok do u recommend anyone else?

  38. anonymous
    • 5 years ago
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    I don't know the particular skill level of people on here. You may have to post and cross fingers.

  39. anonymous
    • 5 years ago
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    Is this an assignment due tomorrow or something?

  40. anonymous
    • 5 years ago
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    ahah yeh you caught me out. due 2pm. i wish i could say im a more dedicated student working til the early hours of the morning doing maths... ah the thrills of life!

  41. anonymous
    • 5 years ago
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    Well, since your timezone's mine, if I see something from you in the morning and I can respond, I will. My day will be hell busy, though.

  42. anonymous
    • 5 years ago
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    Anyways, good luck. I seriously have to hit the sack.

  43. anonymous
    • 5 years ago
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    thanks you are a hero! night!

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